A Bomber On A Military Mission Is Flying Horizontaly At A Height Of 3000m above the ground at 60km/min,it drops a bomb on a target on the ground,determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released
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To determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released, we can use trigonometry.
First, let's consider a right-angled triangle where the vertical line represents the height of the bomber above the ground (3000m) and the horizontal line represents the distance the bomber has traveled (60km/min). The line joining the bomber and the target will be the hypotenuse of this triangle.
Using the Pythagorean theorem, we can find the length of the hypotenuse:
Hypotenuse^2 = Height^2 + Distance^2
Hypotenuse^2 = (3000m)^2 + (60km/min)^2
Now, we need to convert the units so that they are consistent. Converting 60km/min to meters:
60km/min = 60,000m/min
Hypotenuse^2 = (3000m)^2 + (60,000m/min)^2
Simplifying the equation:
Hypotenuse^2 = 9,000,000m^2 + 3,600,000,000m^2/min^2
Hypotenuse^2 = 3,609,000,000m^2 + 3,600,000,000m^2/min^2
Since we are only interested in the acute angle, we can consider the tangent of that angle:
tan(angle) = Height / Distance
tan(angle) = 3000m / 60,000m/min
tan(angle) = 1/20
To find the angle, we take the inverse tangent (arctan) of both sides:
angle = arctan(1/20)
Using a calculator, we find that the angle is approximately 2.86 degrees (rounded to two decimal places).
Therefore, the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released is approximately 2.86 degrees.